Air Squeezed Without Heat

Q. Air at 1 atm fills 8 L of a diesel engine's cylinder. It is compressed so fast that no heat escapes, until its pressure is 32 atm. Taking γ for air as 1.4, what volume does it then fill, in litres? expects 0.673 L

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

Q. Air at 1 atm $P_{0} = 1$ atm fills 8 L $V_{0} = 8$ L of a diesel engine's cylinder. It is compressed so fast that no heat escapes, until its pressure is 32 atm $P_{1} = 32$ atm. Taking γ for air as 1.4 $γ = 1.4$ , what volume does it then fill, in litres?

The playground

Inputs

starting pressure

P_{0}

1 atm

starting volume

V_{0}

8 L

final pressure

P_{1}

32 atm

gamma

γ

1.4

Outputs

final volume

V_{1}

The solution

Solved with AdiabaticProcess

Given

$P_{0} = 1$ atm
$V_{0} = 8$ L
$P_{1} = 32$ atm
$γ = 1.4$

Find $V_{1}$ .

Working

$P_{0} = 1$ atm $= 101325$ Pa
$V_{0} = 8$ L $= 0.008$ m³
$P_{1} = 32$ atm $= 3.2424 \times 10^{6}$ Pa
pv to the gamma is kept(compressed) $P_{0} \cdot {V_{0}}^{γ} = P_{1} \cdot {V_{1}}^{γ}$
$101325 \cdot {0.008}^{1.4} = 3.2424 \times 10^{6} \cdot {V_{1}}^{1.4}$
$V_{1} = 0.673$ L
$V_{1} = 6.72950096 \times 10^{−4}$ m³ $= 0.672950096$ L
$V_{1} \approx 0.673$ Lto 3 significant figures

Conclusion

With no heat let out the gas warms as it is squeezed, so its pressure climbs faster than the volume falls, and a 32-fold rise in pressure needs only a 12-fold fall in volume.

The Ruby

1require_relative "../physics"23question :air_squeezed_without_heat_to_thirty_two_atmospheres do4  kind :numerical5  describes "Air at %{starting_pressure} fills %{starting_volume} of a diesel engine's cylinder. It is compressed so " \6            "fast that no heat escapes, until its pressure is %{final_pressure}. Taking γ for air as %{gamma}, " \7            "%{final_volume}, in litres?"8  given starting_pressure: "1 atm"9  given starting_volume: "8 L"10  given final_pressure: "32 atm"11  given gamma: "1.4"12  asks final_volume: "what volume does it then fill"13end

1require_relative "../questions/air_squeezed_without_heat_to_thirty_two_atmospheres"2require_relative "../science/thermodynamics/adiabatic_process"34solve :air_squeezed_without_heat_to_thirty_two_atmospheres, using: AdiabaticProcess do5  objects :air6  air.goes(:compressed)78  take starting_pressure: air.initial_pressure, starting_volume: air.initial_volume9  take final_pressure: air.final_pressure, gamma: air.ratio_of_specific_heats1011  final_volume = air.final_volume12  answer final_volume:, in: "L", figures: 31314  conclusion "With no heat let out the gas warms as it is squeezed, so its pressure climbs faster than the " \15             "volume falls, and a 32-fold rise in pressure needs only a 12-fold fall in volume."16end

1require_relative "../solutions/air_squeezed_without_heat_to_thirty_two_atmospheres"23play_with :air_squeezed_without_heat_to_thirty_two_atmospheres do4  title "Air Squeezed Without Heat"5end

The laws

quantity	written	units	within	from
pressure	$P$	Pa	0 … ∞	AdiabaticProcess
volume	$V$	m³	0 … ∞	AdiabaticProcess
ratio of specific heats	$γ$	—	—	AdiabaticProcess
starting pressure	$P_{1}$	Pa	0 … ∞	AdiabaticProcess
final pressure	$P_{2}$	Pa	0 … ∞	AdiabaticProcess
starting volume	$V_{1}$	m³	0 … ∞	AdiabaticProcess
final volume	$V_{2}$	m³	0 … ∞	AdiabaticProcess
pressure of air at start of compressed	$P_{0}$	Pa	0 … ∞
volume of air at start of compressed	$V_{0}$	m³	0 … ∞
ratio of specific heats of air	$γ$	—	—
pressure of air at end of compressed	$P_{1}$	Pa	0 … ∞
volume of air at end of compressed	$V_{1}$	m³	0 … ∞

P_{1} \equiv P_{i} (start of motion)

starting pressure

P_{2} \equiv P_{i} (end of motion)

final pressure

V_{1} \equiv V_{i} (start of motion)

starting volume

V_{2} \equiv V_{i} (end of motion)

final volume

P_{0} \cdot {V_{0}}^{γ} = P_{1} \cdot {V_{1}}^{γ}

pv to the gamma is kept(compressed)